Question

Find the value of k such that x=a is a zero of the polynomial x2 -(a+b)x+k. Also, find its other zero.

Answer 1

The gn polynomial is ,

$p(x) = {x}^{2} - (a + b)x + k$
Gn that x=a is a zero of p(x)

By Remainder Theorem

$p(a) = 0$
ie

${a}^{2} - {a}^{2} - ab + k = 0$
ie

$k = ab$
Now the Polynomial is

${x}^{2} - (a + b)x + ab$
Let p(x) be zero

ie
${x}^{2} + ( - a + - b)x + ab = 0$
ie

$(x - a)(x - b) = 0$
Thus the other zero is x=b

Answer 2

x=a
x2-(a+b)x+k
a(2)-(a+b)a+k
let a=1,b=2
1(2)-(1+2)1+k
2-3+k
k=3-2
k=1